Problem
In $△ABC$, suppose $AC>AB$, and if we draw the perpendicular $AD$ from $A$ to $BC$, we have
$$∠DAC>∠DAB \qquad and \qquad DC>DB$$
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Solution
Since $AD⊥BC$, $∠DAB$ and $∠DAC$ are complementary angles of $∠B$ and $∠C$, respectively.
Since $AB<AC$,
$$∠B>∠C,$$
$$∴ \ ∠DAC>∠DAB.$$
Then, if we take $∠DAB’$, which is equal to $∠DAB$, inside $∠DAC$, $B’$ is on $DC$ and $BD=DB’$. Thus,
$$DC>DB’,$$
$$∴ \ DC>DB.$$
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.16